__Question__: Could you explain the concept of a binary star system and how we get its orbit parameters correctly oriented to our view? (Since the system is not likely to be in the plane of our viewing axis initially) Also, how might one find the mass of each, say for a specified binary system?

__Answer__:

Below I sketch a basic binary star system for reference:

The diagram above shows the simplest type of system, with the plane of the binary components A (of mass m(A)) and B (of mass m(B))

*perpendicular*to the viewing plane. This visual binary system, it's important to note, indicates the

*apparent relative orbit*of the 2 stars - since ordinarily (as you pointed out) - the plane of the real orbit will not lie in the plane of the sky, i.e. perpendicular to the line of sight as portrayed. Usually, what we do is observe the motions of the fainter member about the brighter over a period of time sufficient to determine the orbit period, and then obtain the apparent relative orbit.

_{}

^{}

*different eccentricity*, e = c/a. Most importantly, the foci of the original ellipse do not project onto the foci of the projected one. This means the primary (brightest star) though it is located at one focus of the true relative orbit, is not at the focus of the apparent relative orbit.

It is this circumstance which makes it possible to determine the inclination of the

*true orbit*to the plane of the sky. Basically, the problem reduces to finding the angle at which the true relative orbit must be projected in order to account for the amount of displacement of the primary from the focus of the apparent relative orbit.

If the semi-major axis of the true relative orbit (e.g. the one it would have if displayed face-on) has an angular distance of a" (seconds of arc) and if the system is at a distance d parsecs, then the semi-major axis in astronomical units is:

a = (a" x d)

Then the sum of the masses of the two stars is given by Kepler's 3rd law:

m(A) + m(B) = (d a")

^{3}/ P

^{2}

where P is their period.

_{}

^{}

_{}

^{}

We first obtain the total mass from Kepler's 3rd law:

m(A) + m(B) = (d a")

^{3}/ P

^{2 }=

(2.67 pc x 7.5")

^{3}/(50 yr)

^{2}= 3.2

*solar*masses, i.e. 3.2 M

_{s}

So:

m(A) + m(B) = 3.2 M

_{s}

But in terms of

*the center of mass*:

A O

**-----------**-x cm-

**------------------------**o B

where: xB = 2 (xA)

Then:
xB/xA = 2, and: m(A)/m(B) = xB/xA = 2/1

so: m(B) = ½ m(A)

(since the more massive star is always

so: m(B) = ½ m(A)

(since the more massive star is always

*closer to the center of mass*)
Thus,
m(A) » 2.13 M

_{s}, and m(B) » 1.06 M_{s }_{}

_{}

*Spectroscopic binaries*are also of much interest and derive their name because spectroscopic analysis is needed to obtain the radial velocity curve for the system and hence the relative orbital velocity, V, for the pair. The distance around the relative orbit, i.e. its circumference, is just the relative orbital velocity V (deduced from the radial velocity profile) multiplied by the period P. Then the distance between the stars a, is just:

a = (V x P)/ 2π

If, for instance, the relative velocity
is a lower limit, then the separation we obtain is a lower limit for the
system. If this is then applied to Kepler's 3rd law one can obtain a lower
limit to the sum of the masses of the components:

m1 + m2 = a

m1 + m2 = a

^{3}/P^{2}
To fix ideas, say a spectroscopic
binary system is found to have a relative velocity of 100 km/sec and a period
of 17.5 days. Then let component
(1) be 3 times the mass of component (2) and find: a) a
lower limit for the separation of the components, a, and thence a lower limit
to the sum of the masses, and b) the
lower limits on the masses of the components.

First convert 100
km/sec to AU/yr. (Astronomical units per year)

Over one year: t = 3.156 x 10

total distance covered:

Over one year: t = 3.156 x 10

^{7}stotal distance covered:

^{7}s) = 3.156 x 10

^{9}km

But 1 AU = 1.495 x 10

^{8}km

Then, the AU in this total distance:

d/AU = (3.156 x 10

^{9}km)/ (1.495 x 10

^{8}km) = 21.1 AU

The period in yrs. for 17.5 days:

P = 17.5/ (365.25) = 0.048 yr.

Then:

a = (21.1 AU x 0.048 yr/AU)/ 2π = 0.161 AU

The lower limit to the masses is therefore:

m1 + m2 = (0.161)

^{3}/ (0.048)

^{2}= 1.8 solar masses

Since star m1 has
3x the mass of star m2, then: m1
= 3m2

And: m2 + 3m2 =
1.8

**or**4m2 = 1.8
so:

m2 = 1.8/4 =
0.45 solar masses, and

m1 = 3(0.45) = 1.35 solar masses

Basically, in working any binary star problem, say to get the relative masses, one needs to take care the units used are consistent, and also that Kepler's 3rd law is applied in the correct format.

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